package com.hc.programming.linked;

import com.hc.programming.bean.ListNode;

import java.util.Arrays;

/**
 * 给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。
 * <p>
 * 示例 1：
 * 输入：head = [1,2,3,4,5]
 * 输出：[5,4,3,2,1]
 * 示例 2：
 * 输入：head = [1,2]
 * 输出：[2,1]
 * 示例 3：
 * 输入：head = []
 * 输出：[]
 * <p>
 * 提示：
 * 链表中节点的数目范围是 [0, 5000]
 * -5000 <= Node.val <= 5000
 * <p>
 * 进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？
 *
 * @author huangchao E-mail:fengquan8866@163.com
 * @version 创建时间：2024/10/23 20:28
 */
public class 反转链表 {
    public static void main(String[] args) {
        System.out.println(reverseList(new ListNode(Arrays.asList(1,2,3,4,5))));
        System.out.println(reverseList(new ListNode(Arrays.asList(1, 2))));
        System.out.println(reverseList(new ListNode(Arrays.asList())));
    }

    public static ListNode reverseList(ListNode head) {
//        return 递归1(head);
//        return 迭代(head);
        return 递归2(head);
    }

    private static ListNode 递归2(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode newHead = 递归2(head.next);
        head.next.next = head;
        head.next = null;
        return newHead;
    }

    private static ListNode 迭代(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }

    private static ListNode 递归1(ListNode head) {
        if (head == null || head.next == null) return head;

        ListNode newHead = 递归1(head.next);
        ListNode curr = newHead;
        while (curr.next != null) {
            curr = curr.next;
        }
        head.next = null;
        curr.next = head;
        return newHead;
    }
}
